JEE Mathematics Sequence and Series MCQs Set 10

Answer: (c) \( -\frac12 \)

Explanation:
1. Divide each term by the one before it: \( \frac{-12}{24}=-\frac12 \).
2. Check: \( 6\div(-12)=-\frac12 \) and \( -3\div6=-\frac12 \).

Teacher's Note:
1. In a G.P. the ratio of any term to the previous one is constant.
2. A negative ratio makes the terms alternate in sign.

Answer: (a) 1536

Explanation:
1. \( T_n=ar^{n-1} \).
2. \( T_{10}=3\times2^{9}=3\times512=1536 \).

Teacher's Note:
1. The exponent of r is n - 1, not n.
2. 3072 would come from using \( 2^{10} \).

Answer: (b) 6th

Explanation:
1. Here a = 3 and \( r=\frac12 \).
2. \( 3\left(\frac12\right)^{n-1}=\frac{3}{32} \), so \( \left(\frac12\right)^{n-1}=\left(\frac12\right)^5 \).
3. \( n-1=5 \), n = 6.

Teacher's Note:
1. Compare powers of the same base to find n.
2. Check: the terms are 3, 3/2, 3/4, 3/8, 3/16, 3/32.

Answer: (d) -21

Explanation:
1. a = 1, r = -2.
2. \( S_6=\frac{1-(-2)^6}{1-(-2)}=\frac{1-64}{3}=-21 \).

Teacher's Note:
1. Be careful with the sign: \( (-2)^6=+64 \).
2. Check: 1 - 2 + 4 - 8 + 16 - 32 = -21.

Answer: (a) \( na \)

Explanation:
1. When r = 1 every term equals a.
2. The formula \( \frac{a(1-r^n)}{1-r} \) is not defined for r = 1, so we add directly: \( a+a+\dots+a=na \).

Teacher's Note:
1. The standard formula requires \( r\ne1 \).
2. A constant sequence is a G.P. with ratio 1.

Answer: (c) 16

Explanation:
1. \( \frac{T_5}{T_2}=r^3=\frac{81}{24}=\frac{27}{8} \), so \( r=\frac32 \).
2. \( a=\frac{T_2}{r}=24\times\frac23=16 \).

Teacher's Note:
1. Divide the two given terms to eliminate a.
2. Check: 16, 24, 36, 54, 81.

Answer: (c) 9

Explanation:
1. \( 2\cdot2^{n-1}=2^n=512=2^9 \).
2. So n = 9.

Teacher's Note:
1. Write the last term as a power of 2 and compare exponents.
2. The terms are \( 2^1,2^2,\dots,2^9 \).

Answer: (b) \( |r|<1 \)

Explanation:
1. \( r^n\to0 \) as n grows only when \( |r|<1 \).
2. For \( |r|\ge1 \) the partial sums do not settle to one finite value (r = -1 oscillates; r = 1 grows without bound).

Teacher's Note:
1. The condition includes negative ratios between -1 and 1.
2. For r = 1 the sum n a grows without limit.

Answer: (a) 18

Explanation:
1. With four terms in G.P., \( 24=3r^3 \), so \( r^3=8 \) and r = 2.
2. The means are 6 and 12, and \( 6+12=18 \).

Teacher's Note:
1. Inserting 2 means gives 4 terms in total.
2. The sequence is 3, 6, 12, 24.

Answer: (c) 1

Explanation:
1. Three numbers are in G.P. if \( (x+2)^2=x(x+8) \).
2. \( x^2+4x+4=x^2+8x \), so 4x = 4 and x = 1.
3. Terms: 1, 3, 9 with ratio 3.

Teacher's Note:
1. Middle term squared equals the product of the extremes.
2. Check x = 1: 3 squared = 9 = 1 x 9.

Answer: (c) 3280

Explanation:
1. There are 8 terms, a = 1, r = 3.
2. \( S_8=\frac{3^8-1}{3-1}=\frac{6560}{2}=3280 \).

Teacher's Note:
1. The last term is \( 3^7=2187 \) and there are 8 terms.
2. 6560 is the numerator, before dividing by r - 1.

Answer: (d) 3

Explanation:
1. \( a_1=S_1=2 \) and \( a_2=S_2-S_1=8-2=6 \).
2. \( r=\frac{6}{2}=3 \). In general \( a_n=S_n-S_{n-1}=2\cdot3^{n-1} \).

Teacher's Note:
1. The nth term is the difference of consecutive sums.
2. The first term is 2, not 3.

Answer: (b) 512

Explanation:
1. \( a_{10}=S_{10}-S_9=1023-511=512 \).
2. This agrees with \( a_n=2^{n-1} \).

Teacher's Note:
1. 1023 is the sum of ten terms.
2. The terms are 1, 2, 4, 8, ...

Answer: (b) 39366

Explanation:
1. \( r^3=\frac{1458}{54}=27 \), so r = 3.
2. \( a=\frac{54}{27}=2 \).
3. \( T_{10}=2\times3^9=2\times19683=39366 \).

Teacher's Note:
1. Each step of three terms multiplies by 27: 1458 x 27 = 39366.
2. Check: 2, 6, 18, 54, ...

Answer: (d) \( \frac34 \)

Explanation:
1. \( a=\frac12 \) and \( r=\frac13 \).
2. \( S_\infty=\frac{1/2}{2/3}=\frac34 \).

Teacher's Note:
1. Compute 1 - r = 2/3 first.
2. The partial sums 0.5, 0.667, 0.722, ... approach 0.75.

Answer: (a) 5, -10, 20, -40, ...

Explanation:
1. Check ratios of consecutive terms.
2. 5, -10, 20, -40 has ratio -2 throughout.
3. The others have constant difference or non-constant ratios.

Teacher's Note:
1. 1, 3, 5, 7 is an A.P.
2. 1, 4, 9, 16 has ratios 4, 9/4, 16/9, not constant.

Answer: (c) Rs. 13,310

Explanation:
1. The amounts after 1, 2, 3 years form a G.P. with ratio 1.1.
2. \( 10000\times1.1^3=13310 \).

Teacher's Note:
1. Compound growth is geometric, simple interest is arithmetic.
2. Rs. 13,000 would be simple interest.

Answer: (a) a G.P. with common ratio \( \frac12 \)

Explanation:
1. Reciprocals of a G.P. with ratio r have ratio \( \frac1r \).
2. \( \frac{1/4}{1/2}=\frac12 \) and \( \frac{1/8}{1/4}=\frac12 \).

Teacher's Note:
1. Reciprocals of a G.P. stay a G.P.; reciprocals of an A.P. do not stay an A.P.
2. Here r = 2 so the new ratio is 1/2.

Answer: (b) \( \frac23 \)

Explanation:
1. \( \frac{a}{1-r}=12 \), so \( \frac{4}{1-r}=12 \) and \( 1-r=\frac13 \).
2. \( r=\frac23 \).

Teacher's Note:
1. The ratio must be less than 1 in magnitude.
2. Check: \( \frac{4}{1/3}=12 \).

Answer: (b) A.P.

Explanation:
1. \( b^2=ac \) gives \( 2\log b=\log a+\log c \).
2. So the logarithms have a constant difference and form an A.P.

Teacher's Note:
1. Logs turn products into sums, so a G.P. becomes an A.P.
2. The common difference is \( \log r \).